Hi HP,
Thanks for your reply, i can give that in source to select all the files but how do we place them into a directory with name. When i use this format:
jgetfile -j -s "Step 1, job 1:step1_*spool*.txt" -o "%destination%\"
I get error that cant open file.
When i used
jgetfile -j -s "Step 1, job 1:step1_*spool*.txt" -o "%destination%\*spool*.txt"
I get the same error
when i used
jgetfile -j -s "Step 1, job 1:step1_*spool*.txt" -o "%destination%\spool.txt" it only places first spool file and ignore others spool files when the first job has 3 files with names
step1_spool1.txt
step1_spool2.txt
step2_spool3.txt
Please suggest how we can resolve to place all spool files as it is.
Thank you,
Arun